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Thermodynamics question

2001 · Q116
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Thermodynamics question

2001 · Q116

NEETChemistryThermodynamicsMCQ+4 / −1
When 1 mol of gas is heated at constant volume temperature is raised from 298 to 308 K. Heat supplied to the gas is 500 J. Then which statement is correct ?
  1. A
    q = w = 500 J, Δ\DeltaΔE = 0
  2. B
    q = Δ\DeltaΔE = 500 J, w = 0
  3. C
    q = w = 500 J, Δ\DeltaΔE = 0
  4. D
    Δ\DeltaΔE = 0, q = w = −-− 500 J
View written solutionFree

Correct answer: B

As, Δ\Delta ΔE = q + W

Also, W = – pΔ\Delta ΔV

As Δ\Delta ΔV = 0

So, W = 0

⇒\Rightarrow⇒ Δ\Delta ΔE = q

Now, q = 500 J

Thus, Δ\Delta ΔE = q = 500 J

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